2AgNO3 + Zn -> 2Ag + Zn(NO3)2
Put Zn 19g and excessive amount of AgNO3 together in a beaker and start reaction. When Zn is still in a beaker, stop the reaction, and measure the mess of (Zn+Ag). Mass of (Zn+Ag) was 29g. Than what is the mass of Ag in (Zn+Ag)?
reaction
Zn + 2AgNO3.>>>2 Ag + Zn(NO3)2
so for each mole of Zn used, two moles of Ag are produdced
to the ratio of masses Znused/Agproduced=65.4/(2*107l8)
mass Zn left=19-Znused=19-Agproduced(65.4/215)
mass Ag produced=Znused(215/65.4)
Znleft+Agproduced=
19-Agpreduced(65.4/215)+Agproduced=29
solve for Agproduced.
check this reasoning.
To find the mass of Ag in the mixture of (Zn+Ag), we need to subtract the mass of Zn from the total mass of (Zn+Ag). Given that the total mass of (Zn+Ag) is 29g, and we know that Zn has a mass of 19g, we can calculate the mass of Ag.
Mass of Ag = Total mass of (Zn+Ag) - Mass of Zn
Mass of Ag = 29g - 19g
Mass of Ag = 10g
Therefore, the mass of Ag in the (Zn+Ag) mixture is 10g.